Problem 22 - Advanced Engineering Mathematics

Question

If there is a heat transfer from the lateral surface of a thin wire of length \(L\) into a medium at constant temperature \(u_{m}\), then the heat equation takes on the form $$ k \frac{\partial^{2} u}{\partial x^{2}}-h\left(u-u_{m}\right)=\frac{\partial u}{\partial t}, \quad 00 $$ where \(h\) is a constant. Find the temperature \(u(x, t)\) if the initial temperature is a constant \(u_{0}\) throughout and the ends \(x=0\) and \(x=L\) are insulated.

Step-by-Step Solution

Answer

The temperature is determined using separation of variables and solving the resultant spatial and temporal ordinary differential equations.

Step 1: Understand the Problem

We are given a partial differential equation (PDE) to solve for the temperature distribution \(u(x, t)\) in a thin wire with certain conditions. The equation is a heat equation with a source term \(-h(u-u_m)\). The initial temperature is \(u_0\) and the boundaries are insulated, meaning no heat flow through \(x=0\) and \(x=L\).

Step 2: Specify Initial and Boundary Conditions

Initial condition is \(u(x,0) = u_0\) for \(0 < x < L\). The boundary conditions are \(\frac{\partial u}{\partial x}(0, t) = 0\) and \(\frac{\partial u}{\partial x}(L, t) = 0\), indicating insulation at both ends. These will be important for solving the PDE.

Step 3: Simplify the PDE

Recognize that this is a standard form for separation of variables, often simplified by assuming a solution of the form \( u(x, t) = X(x)T(t) \). Substituting this form into the PDE and separating variables will simplify our working expressions.

Step 4: Separate Variables

Substitute \( u(x, t) = X(x)T(t) \) into the PDE to get: \[k \frac{d^2 X}{dx^2}T - h (XT - u_m) = X \frac{dT}{dt}\]Divide the whole equation by \(XT\) to separate the variables:\[\frac{1}{X} k \frac{d^2 X}{dx^2} - \frac{h}{X} (X - u_m) = \frac{1}{T} \frac{dT}{dt}\]This equation suggests that each side must be equal to a constant, \(-\lambda\).

Step 5: Solve the Spatial Part

Solve the spatial part:\[k \frac{d^2 X}{dx^2} + h X = \lambda X\]This is a second order ordinary differential equation (ODE) that can be solved for \(X(x)\) with the boundary conditions \(\frac{dX}{dx}(0) = 0\) and \(\frac{dX}{dx}(L) = 0\).

Step 6: Solve the Temporal Part

Solve the temporal part:\[\frac{dT}{dt} = -\lambda T\]The solution to this ODE is an exponential function \(T(t) = Ce^{-\lambda t}\).

Step 7: Construct General Solution

The general solution for the temperature \(u(x, t)\) will be a linear combination of the solutions of the spatial and temporal parts. It takes the form:\[u(x, t) = \sum_{n=1}^{\infty} (A_n \cos(\sqrt{\frac{\lambda_n}{k}}x) + B_n \sin(\sqrt{\frac{\lambda_n}{k}}x)) e^{-\lambda_n t}\]This accounts for all possible modes that satisfy the PDE and the boundary conditions.

Step 8: Determine Constants from Initial Conditions

Apply the initial condition \(u(x, 0) = u_0\), and use it to find constants by comparing with the form of your solution at \(t = 0\). This typically leads to setting appropriate coefficients that match the initial state of the system.

Key Concepts

Partial Differential Equations

Partial differential equations (PDEs) are equations that involve multiple independent variables, such as time and space, and their partial derivatives. In the heat equation problem discussed, PDEs describe how the temperature distribution in a body changes over space and time. The main objective is to find the function \( u(x, t) \) that represents this temperature distribution.

In our given problem, the PDE is:

• \( k \frac{\partial^2 u}{\partial x^2} - h(u-u_m) = \frac{\partial u}{\partial t} \)

Here, \( u \) is the temperature at point \( x \) and time \( t \), \( k \) is the thermal conductivity, \( h \) is the heat transfer coefficient, and \( u_m \) is the medium temperature. Partial derivatives indicate how the temperature changes with respect to space and time.

PDEs like this one are fundamental for modeling real-world systems that involve continuous changes, such as heat, sound, or fluid flow.

Boundary Conditions

Boundary conditions are crucial for solving a PDE because they provide additional information that helps to uniquely determine the solution. For the heat equation of the thin wire, boundary conditions specify what happens at the edges of the wire. In our problem:

• The ends \(x=0\) and \(x=L\) are insulated, implying \( \frac{\partial u}{\partial x}(0, t) = 0 \) and \( \frac{\partial u}{\partial x}(L, t) = 0 \).
• These conditions mean no heat flows through these ends; essentially, the temperature gradient is zero at these points.

This insulation is modeled by Neumann boundary conditions, which specify the derivative of the solution instead of the values directly. Such specific conditions help make sure that the solution obtained from the PDE is applicable and accurate within the physical scenario described.

Separation of Variables

Separation of variables is a mathematical method used to simplify and solve PDEs by assuming that each variable can be isolated. In our heat equation, we propose a solution of the form \( u(x, t) = X(x)T(t) \), where \( X(x) \) and \( T(t) \) depend only on space and time, respectively.

By substituting this into our PDE and separating based on variables, we translate one complex PDE into two simpler ordinary differential equations (ODEs):

• A spatial equation: \( k \frac{d^2 X}{dx^2} + h X = \lambda X \)
• A temporal equation: \( \frac{dT}{dt} = -\lambda T \)

Here, \(\lambda\) is a constant defined to correlate the separated equations. This technique is powerful because it dismantles a complicated multi-variable problem into more manageable single-variable problems that we can solve individually.

Finally, we combine these solutions for \( X(x) \) and \( T(t) \) to get the complete solution for the PDE.

Temperature Distribution

The main goal of solving the heat equation is to determine the temperature distribution, \( u(x, t) \), in the wire over time. Once we apply the separation of variables and solve the ODEs for \( X \) and \( T \), we express the general solution for \( u(x, t) \) as a series:

• \( u(x, t) = \sum_{n=1}^{\infty} (A_n \cos(\sqrt{\frac{\lambda_n}{k}}x) + B_n \sin(\sqrt{\frac{\lambda_n}{k}}x)) e^{-\lambda_n t} \)

This series form includes various "modes" of temperature change shaped by the spatial and temporal components.

The solution is an infinite sum of such modes, each scaled by coefficients \(A_n\) and \(B_n\). These coefficients are determined using initial conditions, here by setting \( u(x, 0) = u_0 \). Once you incorporate the boundary and initial conditions, you can accurately describe how temperature varies throughout the wire over time.